@floyd617 i leave those terms intact because it provides a useful expression: 2/n^2 and the series associated with that converges. you have to realize that you are trying to show the given expression is smaller than some convergent series (or larger than some divergent series). it is not mechanical to do this: how can one justify it? for this example, i found a way! (i hope i am remember correctly what is in the video)
For those wondering about the 2/(n^2) expression. Consider that the expanded terms he showed are constantly multiplying by something *smaller than one*; thus, the sum would constantly be getting smaller. If you remove these terms making your sum smaller, and just keep the final two terms, you will then have a series that is in fact larger than the initial series given. He also keeps the final two because together they make up a p-series, which is easy to tell divergence/convergence.
@AznVamp24 if you use the direct comparison test and show that the sum of the series is smaller than the sum of the series associated with 1/n you have not shown anything as that latter series diverges.
it seems to me like the last example would be much better solved by the ratio test, however this method explains the true fundamentals of the comparison test, so good job
These videos are so helpful, thanks so much. I have my final tomorrow and I've been trying to study the book, but having someone explain it makes it so much easier.
thank you for all your help. ever since high school i've been watching you're videos and now I've passed my first year of math and i am finally FREE from it. thanks again!
i think in order for you to use the comparison test directly to show your series is div, the thing being compared must also be div (like example two) BUT it must also be a lower limit (not an upper limit like in patrickJMT's 2nd ex).
i agree, you can just look at it, realize it is close to being a p-series, and so say it diverges. you can say that only because of the limit comparison test (that is the justification); otherwise, you are just 'hand waving'
@mikegovikes it would be, but the point of this vid is to teach the limit and direct comparison tests. For alot of sums you can choose from almost any test to use on it, however what decides which test to use is basically a) preference b) ease of use c) and in some cases, like geometric series, the fact that you can actually define the sum of a series
An example where both the numerator and denominator are constants raised to "n" would be great! ^_^ thanks for the videos, it really helped me understand because i can rewind and watch again until i know what you're talking about.
why i didnt find yout videos earlier ??? whyyyy ? great lessons i am giving tommorow morning and now i am gonig to see as many of your videos as i can till 2mmorow
I think the issue is that video 122 in the Calc 2 playlist is a bit more of an intro to these ideas versus this video (121 in the playlist). So this one just uses a few terms such as p-series in a way that isn't immediately obvious.
why did u use limit comparison test in the second one instead of direct comparison... i get that an is bigger than bn, and because of the p series it was already divergent, i could have used direct comparison ... ?
Also, 1 other question. How did you conclude the 2 / n^2 in the next problem? I see how you got it, but why are you able to do that? Is there another (better) series test to tackle this problem with? It's just hard for me to wrap my head around the 2 / n^2 part.
@patrickJMT At the beginning, you said: [1 + sin(n)] / [10^n] is less than or equal to 2/10^n, "for all n is greater than or equal to 1". Why did you choose 1? Isn't it 0, because that's what our original series had n equal to? Thank you for your help! Your videos are great!
@pochankitty not free. dont you see the commercials? lol. great teacher nonetheless. definitely always rely on you, patrickJMT, when my teachers fail to teach
Sir, could you please answer me these questions : 1) You figure out the bsubn series by figuring out the highest powers in both the numerator and the denominator? 2) Suppose i'm given a question and asked to find whether it converges or diverges, can i just ignore the direct comparison and do the limit comparison test? cuz the limit test seems easier to me than the direct one. 3) for a p series, can we conclude that the series diverges if p< 1? Thanks in advance.
your video is so helpful for me to understand the concept！But may i want to know is there any differences between comparison and limit comparison？I cannot find any difference in the three examples
I hope you make a decent living off of these videos and the apps (I bought many of them, myself!). They are so unbelievably helpful and have saved my grade on many occasions! Thank you :D
For the problem at 3:13, why don`t you say that the series is divergent if its bigger than the divergent p-series. Why use the limit comparison test instead?
according to the p-series, 1/n^1/2 is divergent. since the limit of (sqrt(n)/n-1)/1/n^1/2 is 1, which is finite and >0, according to the limit comparison test, since the former is divergent, the latter is also divergent. you can also use divergence test since the limit of sqrt(n)/n-1 is not equal zero, it thus diverges.
+Seandt Calculus -_- in the class room we just say omg hehehe but here 2 hours am done from all the chapter :3 i wish i've revised here to be the hero there :3
+samrath singh I tried the problem using the ratio test and got a solid answer. When you use the LCT or DCT, then this becomes a shaky argument. I confirmed with my professor, and he said this is a ratio test argument. Btw I did get convergent after using the ratio test. Hope this helps!
No, if you took (1/n), then the comparison test would be inconclusive. Since (1/n) > (n! / n^n) , if Sum(1/n) diverges then you CANNOT conclude anything about (n! / n^n). In which case, you need to then pick a different sequence to compare to. When you pick (2/n^2), btw you couldve picked any constant divided by n^2, since (2/n^2) > (n! / n^n) AND (2/n^2) converges, then (n! / n^n) MUST also converge. Picking the harmonic series leaves with you an inconclusive argument. This happens often with the comparison test.
@stealinglemons I do not think he covered P-Series. I had to look it up in my book and get a quick definition of it. P series is refering to the exponent of the 1 / n^p. If p > 1 ; Converges if p
Omg! I did not understand why only (2/n) and (1/n) were used until now! My math teacher showed this example a month ago and I remember something being negative ^^". Either my memory sucks or that explanation failed, but I wasn't the only one extremely confused that day.
when do you know to use limit comparison or direct comparison? do you just use limit comparison when you cant do direct? would you be able to do limit comparison instead of direct and get the same answer? lets say if i did limit comparison for your first example. THANKS ALOT!
what made you pick out the last two instead of the last one? because if you only used 1/n, it would be divergent because its harmonic series but you chose the last two making it a p-series?
I am a little confused on that part as well but my guess would be: n is a really big number. Even if you include the 3/n which would be 3/n*2/n*1/n= 6/n^3 this is still a convergent p series, so the answer comes out to be the same.
I understand how sin(n) is always -1 < sin(n) < 1 .... so I thought it turned into (-1)^n. How do you simply come up with 2 / 10^n as your b-sub-n? Would you mind being at least slightly detailed? I'm just a little confused. Thank you in advance. I hope you have the time to respond. Your videos are excellent and I am extremely appreciative of the work you do.
think about what you said... n^(1/2) is the square root of n. consider n=4. 4^(1/2) = 2. 2*2 = 4. not 4^(1/4). Multiplying numbers with exponents is a slightly different rule set than multiplying fractions.
Yeah, a lot of people are confused about the last example where you compare the original to (2/n)(1/n). What's stopping you from just comparing it to only (1/n)? And why are those valid choices since they're both less than 1 as well? Thank you! :]
yeah, I did! The only thing that is challenging for me in Calc is the intermediate Algebra. I would rather screw up on the simple stuff than not have the intellectual capacity for the fun stuff. I hope that is not to complicated for you?!
I am confused as to why you left out the 3/n but not the 2/n and 1/n ? I see how the terms you left out are less than one, but why not 2/n and 1/n? Maybe this is a silly question, but I'm not seeing it for some reason
thanks for ur videos, when u say p series, do u mean power series? I'm a bit confused as I'm thinking of power series as things in the form of 1/(1-x)..am I wrong? thanks..
wait, in the last example, why do you keep the last two terms: 2/n and 1/n ? i mean, how do you know not to just keep the last term 1/n? (which would then make it divergent)
@patrickJMT so is it ok if i use direct comparison for this question and not going the extra mile and do limit comparison?i still don't get what do you mean by 'hand waving'....
I understand we compared it to 1/n which dv. Im just a little confused about all the test and what it needs to be greater or less then in order for it to cv or dv
I never understand how the b sub n series is derived. Is it embedded in the problem itself or do you have to do something to the given series that will enable you to obtain the b sub n series.
that's just because infinity isn't a number at all. It's just a concept used to understand that there is no largest number. 1^anything is 1 though... That's really the point of using limits if you're going to use infinity. go to wolframalpha and put this in: lim as n approaches infinity 1^n
could we use direct comparison test in ex 2: bn=1/(n^1/2) is divergent an> or equal to bn (n)^1/2 / n-1 > or equal to ( 1/n^1/2) since bn is divergence an is divergenc right? Thank you alot
Because being less than a divergent series means nothing, but being less than a convergent series actual helps us. An, Bn = a sub n and b sub n; respectively For 0
![Comparison Tests [ Direct and Limit]](http://i.ytimg.com/vi/ITAUERlNDrU/mqdefault.jpg)
@floyd617 i leave those terms intact because it provides a useful expression: 2/n^2 and the series associated with that converges. you have to realize that you are trying to show the given expression is smaller than some convergent series (or larger than some divergent series). it is not mechanical to do this: how can one justify it? for this example, i found a way!
(i hope i am remember correctly what is in the video)
For those wondering about the 2/(n^2) expression.
Consider that the expanded terms he showed are constantly multiplying by something *smaller than one*; thus, the sum would constantly be getting smaller. If you remove these terms making your sum smaller, and just keep the final two terms, you will then have a series that is in fact larger than the initial series given.
He also keeps the final two because together they make up a p-series, which is easy to tell divergence/convergence.
@AznVamp24 if you use the direct comparison test and show that the sum of the series is smaller than the sum of the series associated with 1/n you have not shown anything as that latter series diverges.
Your videos are the only ones that can actually seem to make some sense out of this for me. Thank you thank you thank you for making them!
it's so refreshing seeing a fellow left-hander teach math. this was a very helpful video. thanks.
much better than my teachers lessons in class! I can use your videos with the CK12 calc book to make my understanding much better, thanks :)
it seems to me like the last example would be much better solved by the ratio test, however this method explains the true fundamentals of the comparison test, so good job
These videos are so helpful, thanks so much. I have my final tomorrow and I've been trying to study the book, but having someone explain it makes it so much easier.
thank you for all your help. ever since high school i've been watching you're videos and now I've passed my first year of math and i am finally FREE from it. thanks again!
i think in order for you to use the comparison test directly to show your series is div, the thing being compared must also be div (like example two) BUT it must also be a lower limit (not an upper limit like in patrickJMT's 2nd ex).
Once again I get stuck on a problem and magically ur video is on the same one! Thanks sir
You are so much more clear than my teacher, thank you!
i agree, you can just look at it, realize it is close to being a p-series, and so say it diverges. you can say that only because of the limit comparison test (that is the justification); otherwise, you are just 'hand waving'
patrickJMT is the god of calculus thank you so much for the help
Thank you so very much. My calculus teacher is out of control. Thank you so much.
@mikegovikes it would be, but the point of this vid is to teach the limit and direct comparison tests. For alot of sums you can choose from almost any test to use on it, however what decides which test to use is basically
a) preference
b) ease of use
c) and in some cases, like geometric series, the fact that you can actually define the sum of a series
omg you explain these things better than my professor. your tutorials are awesome! :)
ty patrickjmt
you just saved me 1 hour of reading through the difficult language of my calc book.
I really enjoy your calm voice
An example where both the numerator and denominator are constants raised to "n" would be great! ^_^
thanks for the videos, it really helped me understand because i can rewind and watch again until i know what you're talking about.
why i didnt find yout videos earlier ??? whyyyy ?
great lessons i am giving tommorow morning and now i am gonig to see as many of your videos as i can till 2mmorow
@Payme4ril p-series
1/n^c
with c being a constant. if c > 1 it converges
if c
@eternity23211 part of the fun is figuring that out
I think the issue is that video 122 in the Calc 2 playlist is a bit more of an intro to these ideas versus this video (121 in the playlist). So this one just uses a few terms such as p-series in a way that isn't immediately obvious.
why did u use limit comparison test in the second one instead of direct comparison... i get that an is bigger than bn, and because of the p series it was already divergent, i could have used direct comparison ... ?
YOu rock! thanks so much for the help, I'm studying civil engineering right now and this is really good stuff because it helps.
Again, thanks so much.
When multiplying the same base with different powers, add the powers.
IE: 2^(1/4)*2^(3/4)=2
Also, 1 other question. How did you conclude the 2 / n^2 in the next problem? I see how you got it, but why are you able to do that? Is there another (better) series test to tackle this problem with? It's just hard for me to wrap my head around the 2 / n^2 part.
that third one though. haha I don't know how I would've thought of that on my own
+kalef1234 3rd u need to use lembert directly while us ee n! and a^n
For that problem, I think it would have been easier to use the ratio test
ratio test is really useful for factorials.
andy brisman I tested it, the ratio test is inconclusive for that problem
@@carl7018 I have reached to (n/n+1)^(n+1) then putting infinity causes to inconclusion. Comparation test is the only way i think.
@patrickJMT At the beginning, you said:
[1 + sin(n)] / [10^n] is less than or equal to 2/10^n, "for all n is greater than or equal to 1". Why did you choose 1? Isn't it 0, because that's what our original series had n equal to?
Thank you for your help! Your videos are great!
@AznVamp24 1/n is a harmonic series which is divergent...but 2/n^2 is convergent
(n^(1/2))/n^1 = 1/n^(1/2) ; u are just subtracting exponents
@pochankitty not free. dont you see the commercials? lol. great teacher nonetheless. definitely always rely on you, patrickJMT, when my teachers fail to teach
Sir, could you please answer me these questions :
1) You figure out the bsubn series by figuring out the highest powers in both the numerator and the denominator?
2) Suppose i'm given a question and asked to find whether it converges or diverges, can i just ignore the direct comparison and do the limit comparison test? cuz the limit test seems easier to me than the direct one.
3) for a p series, can we conclude that the series diverges if p< 1?
Thanks in advance.
your video is so helpful for me to understand the concept！But may i want to know is there any differences between comparison and limit comparison？I cannot find any difference in the three examples
the second problem you did, wouldn’t it be easier if you just said that since an>bn and bn is divergent, then an will also be divergent?
Why did you just randomly pick 2 for 2/10^n ? at around 1:00
I hope you make a decent living off of these videos and the apps (I bought many of them, myself!). They are so unbelievably helpful and have saved my grade on many occasions! Thank you :D
have fun :)
you are da best. you should do videos on physics too!
For the problem at 3:13, why don`t you say that the series is divergent if its bigger than the divergent p-series. Why use the limit comparison test instead?
according to the p-series, 1/n^1/2 is divergent. since the limit of (sqrt(n)/n-1)/1/n^1/2 is 1, which is finite and >0, according to the limit comparison test, since the former is divergent, the latter is also divergent. you can also use divergence test since the limit of sqrt(n)/n-1 is not equal zero, it thus diverges.
Why do "professors" make this so confusing?
In awe of a Brilliant Mind
+Seandt Calculus -_- in the class room we just say omg hehehe but here 2 hours am done from all the chapter :3 i wish i've revised here to be the hero there :3
YOU ARE THE BEST !!
thank you so much, this is much more helpful than lecture
Thattutorguy is freaking overrunning every math tutor channel.....
In third problem why you took 2 terms, if we only take last 1/n then it would be harmonic and divergent ?
+samrath singh I tried the problem using the ratio test and got a solid answer. When you use the LCT or DCT, then this becomes a shaky argument. I confirmed with my professor, and he said this is a ratio test argument. Btw I did get convergent after using the ratio test. Hope this helps!
No, if you took (1/n), then the comparison test would be inconclusive. Since (1/n) > (n! / n^n) , if Sum(1/n) diverges then you CANNOT conclude anything about (n! / n^n). In which case, you need to then pick a different sequence to compare to. When you pick (2/n^2), btw you couldve picked any constant divided by n^2, since (2/n^2) > (n! / n^n) AND (2/n^2) converges, then (n! / n^n) MUST also converge. Picking the harmonic series leaves with you an inconclusive argument. This happens often with the comparison test.
FXXX YEA. IAM going to pass my CACL Test, THANKS MAN!
Very helpful. Keep it up, I got a calc test coming up :P
@stealinglemons I do not think he covered P-Series. I had to look it up in my book and get a quick definition of it.
P series is refering to the exponent of the 1 / n^p. If p > 1 ; Converges if p
@patrickJMT thanks
I think I get it now. You have to test An for the 2nd example because if the limit of An/Bn happened to be 0 it would converge?
do you have a video that explains the concept of the direct comparison and integral comparison tests, without just jumping straight into examples?
Omg! I did not understand why only (2/n) and (1/n) were used until now! My math teacher showed this example a month ago and I remember something being negative ^^". Either my memory sucks or that explanation failed, but I wasn't the only one extremely confused that day.
when do you know to use limit comparison or direct comparison? do you just use limit comparison when you cant do direct?
would you be able to do limit comparison instead of direct and get the same answer? lets say if i did limit comparison for your first example.
THANKS ALOT!
what made you pick out the last two instead of the last one? because if you only used 1/n, it would be divergent because its harmonic series but you chose the last two making it a p-series?
@khilozozo02 what dont you get
I am a little confused on that part as well but my guess would be: n is a really big number. Even if you include the 3/n which would be 3/n*2/n*1/n= 6/n^3 this is still a convergent p series, so the answer comes out to be the same.
I understand how sin(n) is always -1 < sin(n) < 1 .... so I thought it turned into (-1)^n. How do you simply come up with 2 / 10^n as your b-sub-n? Would you mind being at least slightly detailed? I'm just a little confused. Thank you in advance. I hope you have the time to respond. Your videos are excellent and I am extremely appreciative of the work you do.
on the second example, when you kept the last two terms, why did you keep only the last two?
Not sure what we are looking for when doing either comparison test? Something that looks like a geometric or p-series?
it'd also work since 6/n^3 is also a convergent p-series
There are 2 many tests , how to know which one to use ?
think about what you said... n^(1/2) is the square root of n. consider n=4. 4^(1/2) = 2.
2*2 = 4. not 4^(1/4).
Multiplying numbers with exponents is a slightly different rule set than multiplying fractions.
@mccleery26 uh, try again.
Yeah, a lot of people are confused about the last example where you compare the original to (2/n)(1/n).
What's stopping you from just comparing it to only (1/n)? And why are those valid choices since they're both less than 1 as well?
Thank you! :]
if you wanted to, could you use (3/n)? or is it only (2/n &1/n)?
yeah, I did! The only thing that is challenging for me in Calc is the intermediate Algebra. I would rather screw up on the simple stuff than not have the intellectual capacity for the fun stuff. I hope that is not to complicated for you?!
for your 2nd example you compared the original series to 1/n^1/2 which is div...so why the go the extra mile and do the limit comp?
Thanks for the vid, the part i'm stuck on is how do u know if the new series you make is supposed to be bigger or smaller than the original.
patrick, on 6:55 why did you choose to keep 2/n & 1/n to compare to n!/n^2?
would the ratio test also be useful for the last problem: n!/n^n ?
I am confused as to why you left out the 3/n but not the 2/n and 1/n ? I see how the terms you left out are less than one, but why not 2/n and 1/n? Maybe this is a silly question, but I'm not seeing it for some reason
thanks for ur videos, when u say p series, do u mean power series? I'm a bit confused as I'm thinking of power series as things in the form of 1/(1-x)..am I wrong? thanks..
wait, in the last example, why do you keep the last two terms:
2/n and 1/n ?
i mean, how do you know not to just keep the last term 1/n? (which would then make it divergent)
@patrickJMT so is it ok if i use direct comparison for this question and not going the extra mile and do limit comparison?i still don't get what do you mean by 'hand waving'....
@patrickJMT So are you saying that I could have used the last THREE terms as well? That would have still proved the exact same thing, no?
in questions like the last question do we always keep the last terms?
yeah he does
Is the direct comparison test the same as the basic comparison test
Was there ever a video on p-series?
I understand we compared it to 1/n which dv. Im just a little confused about all the test and what it needs to be greater or less then in order for it to cv or dv
is this the only test you can do for the second example? or could you do the ratio test aswell?
shouldn't it be for all n greater than or equal to 0 since the summation goes from 0 to infinity?
Thx u Patrickjmt
I never understand how the b sub n series is derived. Is it embedded in the problem itself or do you have to do something to the given series that will enable you to obtain the b sub n series.
What about 1/n! from zero to infinity
Thanks!
thank you!
hey Patrick why did we use LCT on the second example. why didn't we use comparison test. I'm confused when do we use LCT and Whe do we use CT.
I love how you're trying not to ruin the surprise of the second example. Chortle.
great video.
May I ask u something?
At the part around 3:10, u wrote root n/n = 1/n power half
I still don't understand how u got that
Thanks :)
@mccleery26 add the exponents when multiplying and you get n^1...
that's just because infinity isn't a number at all. It's just a concept used to understand that there is no largest number. 1^anything is 1 though... That's really the point of using limits if you're going to use infinity. go to wolframalpha
and put this in: lim as n approaches infinity 1^n
could we use direct comparison test in ex 2:
bn=1/(n^1/2) is divergent
an> or equal to bn
(n)^1/2 / n-1 > or equal to ( 1/n^1/2)
since bn is divergence
an is divergenc
right?
Thank you alot
+Rose 201 I had the same thought process. Did you ever find out if you were correct?
AA A I’m pretty sure you can
Because being less than a divergent series means nothing, but being less than a convergent series actual helps us.
An, Bn = a sub n and b sub n; respectively
For 0
the original series is not a p-series though
on example 1, why does 1/10 have to be less than 1? shouldn't it be 1/(10^n) less than 1/(n^2)?
Good eye there xSilver.....
I have to agree with my math theacher... I'm not smart, just stubborn.
Doesn't the limit comparison only work for positive term series?